Groups, order G = 60, G simple

3. We make the assumption that G is not isomorphic to A5
Then "given G cannot have a subgroup of index 2, 3, 4, 5," I can
get the result.
My problem is I don't know why the quoted statement is true.
Clear for 2 alright, given G simple, but 3, 4, 5?
Is this a fairly obvious result (if so, a hint please), or is it difficult to
prove and should i move on until I know more about groups?

Some of them are clear: if there is a subgroup of index d, then there is a homomorphism to S_d, the symmetric group on d elements - i.e. the (transitive, i.e. there is only one orbit) group action on the cosets. So for d=2,3,4, the orders of S_d are 2,6,24, so it is not injective, hence has a kernel, which would be normal. This only leaves the index 5 case. Again, this must be injective, so it is up to you to figure out what the (transitive) subgroups of order 60 are in S_5. There of course might be a better way to do the question than this.

A little "over my head". Not your fault though, just haven't heard of the terms "transitive subgroup" and "transitive group action". Unfamiliar with opening result as well, actually thought it was Cayley's theorem at first. Really appreciate the help, now happy in the knowledge that I wasn't overlooking some result from my course. Will leave topic until I know more about groups.